Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 5 / ii / a

For a positive integer $k$, cancellation of $U^{†}U=I$ gives $A^k=U{D}^k{U}^{†}$. The trace is invariant under matrix similarity, since $\mathrm{tr}(XY)=\mathrm{tr}(YX)$, and the determinant is multiplicative. ThereforeThis includes zero eigenvalues without requiring $A$ to be invertible.